Is it true that there is an irreducible real representation $(\rho, V)$
of $\mathfrak g$ such that $\beta$ is a weight of $V$ and there is a nonzero
vector $v\in V_{\beta}$ annihilated by $\mathfrak u_{\mathfrak p}$? I.e.
$\rho(u)v=0$ for every $u \in \mathfrak u_{\mathfrak p}$.

In the case where $\mathfrak g$ is split and $\mathfrak p$ is the minimal parabolic subalgebra we get an affirmative answer from highest weight theorem.
In this sense what I am asking can be considered as a generalization of
highest weight theorem.

It would also be interesting to have an answer in the case where $\mathfrak g $
is split. Maybe experts in Lie algebra can answer this question immediately

1 Answer
1

You are asking for a "real" representation. That is not possible in general. For example, take $G=SU(2,1)$ as the group preserving the Hermitian form given by
$$\begin{pmatrix} 0& 0 & 1\cr 0 & 1 & 0 \cr 1 & 0 & 0\end{pmatrix}.$$
and take the standard representation on ${\mathbb C}^3$. The highest weight is the weight of $e_1$ (the first basis vector). However, this representation is not defined over ${\mathbb R}$, since that would mean that $SU(2,1)$ can be conjugated into $SL(3,{\mathbb R})$.

If you drop the "real" assumption, then I think it is indeed possible to find a representation.

$\begingroup$In this case I will consider $\mathbb C^3$ as $\mathbb R^6$. As in my question, I am considering real Lie algebras and relative root systems. So the group $G=SU(2, 1)$ has rank one.$\endgroup$
– ronggangMar 3 '15 at 5:53

$\begingroup$Of course, then the representation is not absolutely irreducible, and the highest weight space over the reals, is two dimensional$\endgroup$
– VenkataramanaMar 3 '15 at 6:10